The standard counterexample is the field $F$ of finitely-tailed Laurent series over $\mathbb{R}$: series of the form $\sum a_n t^n$ where $a_n \in \mathbb{R}$, exponents $n \in \mathbb{Z}$, but with only finitely many negative exponents (i.e., $a_n = 0$ for all but finitely many negative integers $n$.)
Addition and multiplication are defined just like for power series; you should verify that the "finitely many negative exponents" condition is essential to ensuring that multiplication of finitely-tailed Laurent series is well-defined. Check that non-zero elements of $F$ have multiplicative inverses: a series that starts with $a_n t^n$ has an inverse that starts with $a_n^{-1} t^{-n}$. (Here, and below, when I say that a nonzero element of $F$ starts with $a_n t^n$, I mean that $n$ is the lowest integer for which $a_n \ne 0$.)
Ordering is defined as follows: An element is positive if its starting term has positive coefficient. It's easy to check: (1) for every nonzero $x \in F$, exactly one of $x$ and $-x$ is positive; (2) the sum and product of 2 positive elements is positive.
So $F$ is an ordered field. Note that $F$ is not Archimedean: $t^{-1}$ is bigger than every integer. The field $F$ also does not satisfy the least upper bound property: $t^{-1}$ is an upper bound for the integers, but there is no least upper bound (check that if $x$ is such an upper bound, then $x/2$ is a smaller one.)
To discuss Cauchy completeness, we have to put a metric on $F$, as follows. Pick any positive real number $q>1$. If a nonzero element $x \in F$ starts with $a_n t^n$, then define $|x| = q^{-n}$ (and of course define $|0|=0$), and then define the distance between $x$ and $y$ to be $|x-y|$. Check that this is a metric space. (Caution: This doesn't extend the existing metric on $\mathbb{R}$; all real numbers have the same size under this metric.) In fact, it makes $F$ into a valued field; we have $|xy| = |x||y|$.
This is a "non-Archimedean" metric because it satisfies the strong triangle inequality $d(x,z) \le \max(d(x,y), d(y,z))$. For a non-Archimedean metric, a sequence is Cauchy iff $|x_n - x_{n+1}| \to 0$ as $n \to \infty$.
To prove that $F$ is complete under this metric, note that for a Cauchy sequence in $F$, the coefficients for a fixed exponent $t^n$ must be eventually constant; otherwise the difference between consecutive terms could never be smaller than $q^{-n}$. Call this constant $a_n$; then the Cauchy sequence converges to $\sum a_n t^n$.
Suppose that $\langle x,y\rangle\in (X\times Y)\setminus G_f$. Then $y\ne f(x)$, and $Y$ is Hausdorff, so there are disjoint open $U,V$ in $Y$ such that $y\in U$ and $f(x)\in V$. Since $f$ is continuous, there is an open nbhd $W$ of $x$ such that $f[W]\subseteq V$; clearly $W\times U$ is an open nbhd of $\langle x,y\rangle$ disjoint from $G_f$.
It is necessary to require that $Y$ be Hausdorff. For a simple example, let $X=\{0,1\}$ have the discrete topology, and let $Y=\{0,1\}$ with the Sierpiński topology, whose open sets are $\varnothing,\{0\}$, and $Y$ itself. Let $f:X\to Y$ be the identity function; $f$ is certainly continuous, since $X$ is discrete, but $\langle 0,1\rangle$ is in the closure of $G_f$, since every nbhd of $\langle 0,1\rangle$ contains $\langle 0,0\rangle$.
Added: The space $Y$ in that example is $T_0$ but not $T_1$; here’s an example in which $Y$ is $T_1$. Let $X=\mathbb{N}\cup\{p\}$, where $p\notin\mathbb{N}$, and let $Y=\mathbb{N}\cup\{p,q\}$, where $q\notin\mathbb{N}$ and $p\ne q$. In both $X$ and $Y$ the points of $\mathbb{N}$ are isolated, and in both $X$ and $Y$ a local base at $p$ consists of all sets of the form $\{p\}\cup(\mathbb{N}\setminus F)$ such that $F$ is a finite subset of $\mathbb{N}$. Finally, a local base at $q$ in $Y$ consists of all sets of the form $\{q\}\cup(\mathbb{N}\setminus F)$ such that $F$ is a finite subset of $\mathbb{N}$. The points $p$ and $q$ in $Y$ do not have disjoint open nbhds; they are the only pair of points in $Y$ that cannot be separated by disjoint open sets.
Let $f:X\to Y:x\mapsto x$ be the identity function; it’s easy to see that $f$ is not just continuous, but an embedding. The point $\langle p,q\rangle\in X\times Y$ is not in $G_f$, but you can check that if $U$ is an open nbhd of $\langle p,q\rangle$ in $X\times Y$, then there is an $m\in\mathbb{N}$ such that $\langle n,n\rangle\in U$ whenever $n\ge m$, so $U\cap G_f\ne\varnothing$. Thus, $\langle p,q\rangle$ is in the closure of $G_f$.
Best Answer
"$\Leftarrow$": Short version: Below is a proof that MCT implies the LUB property. The basic idea is to take the set $U$ of upper bounds of a nonempty set $A$ that is bounded above, and to construct an increasing sequence $(x_k)$ with no term in $U$, such that $x_k+\varepsilon_k$ is in $U$ for all $k$, and $\varepsilon_k\to 0$. MCT implies that $(x_k)$ converges, and the limit can be shown to be $\sup A$. If you are only looking for hints, read no further.
Proof by contradiction: If $\mathbb{N}$ is bounded, then the sequence $(n)_n=(1,2,3,4,\ldots)$ converges by hypothesis to some real number $x$, and then $(n+1)_n$ converges to $x$ and to $x+1$, a contradiction.$\square$
Let $A$ be a nonempty set of real numbers that is bounded above, and let $U$ be the set of upper bounds of $A$.
Claim 1: For all $\varepsilon>0$, $U-\varepsilon:=\{u-\varepsilon:u\in U\}$ is not contained in $U$.
Proof by contradiction: Let $\varepsilon>0$. If $U-\varepsilon\subseteq U$, then $U-n\varepsilon\subseteq U$ for all $n$. There exists $u\in U$, and $U$ therefore contains the ray $\{x:x>u\}$. This implies by the Archimedean property that $\mathbb R=\bigcup\limits_{n=1}^\infty (U-n\varepsilon)\subseteq U$, which contradicts $A\neq\emptyset$.$\square$
Claim 2: $\bigcap\limits_{n=1}^\infty\left(U-\frac{1}{n}\right)\subseteq U$.
Proof: Suppose that $x$ is in $\bigcap\limits_{n=1}^\infty\left(U-\frac{1}{n}\right)$, and that $y>x$. By the Archimedean property there exists $n\in\mathbb N$ such that $x+\frac{1}{n}<y$. But $x+\frac{1}{n}$ is in $U$, so this implies that $y$ is not in $A$. Since $y$ was arbitrary, $x$ is an upper bound for $A$.$\square$
Let $x_1$ be an element of $(U-1)\setminus U$ (which is nonempty by Claim 1). Since $x_1$ is not in $U$, by Claim 2 there exists $n_1>1$ such that $x_1$ is not in $U-\frac{1}{n_1}$. Let $x_2$ be an element of $\left(U-\frac{1}{n_1}\right)\setminus U$. There exists $n_2>n_1$ such that $x_2$ is not in $U-\frac{1}{n_2}$, and we can take $x_3\in \left(U-\frac{1}{n_2}\right)\setminus U$; and so on. This yields an increasing sequence of positive integers $1=n_0<n_1<n_2<n_3<\cdots$ and an increasing sequence of real numbers $(x_k)$ such that $x_k$ is in $\left(U-\frac{1}{n_{k-1}}\right)\setminus U$. Then $(x_k)$ is bounded above by each element of $U$. Assuming MCT, $(x_k)$ converges to some real number $x$. Since $(x_k)$ is bounded above by each element of $U$ and $\lim x_k=x$, $x$ is less than or equal to each element of $U$. Since $x\geq x_k$ for all $k$, $x$ is in $\bigcap\limits_{j=1}^\infty\left(U-\frac{1}{n_j}\right)=\bigcap\limits_{n=1}^\infty\left(U-\frac{1}{n}\right)\subseteq U$. Therefore $x$ is the smallest element of $U$, which means $x=\sup A$.$\square$